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The angular deviation of $5^{t h}$ order dark fringe is $12^{0}$ in a single slit experiment. If the width of slit is $9 μ m$ then the wave length of the incident light is__

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$4862 A^{0}$

$5892 A^{0}$

$6022 A^{0}$

$3768 A^{0}$

answer is D.

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Detailed Solution

$θ = 12^{0}$

$d = 9 μ m 12 (\frac{π}{180}) = \frac{5 λ}{d}$

$λ = \frac{12 π d}{5 \times 180} = \frac{12 \times 3.14 \times 9 \times 10^{- 6}}{900} = 3768 A^{0}$

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